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Question

A mass m is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. The whole system is at rest. At t=0, m is released and the system falls freely under gravity. Assume that natural length of the spring is L0, its initial stretched length is L and the acceleration due to gravity is g. What is distance between masses as a function of time ?

A
L0+(LL0)cos2kmt
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B
L0+(LL0)coskmt
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C
L02+(L+2L0)cos2kmt
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D
L0+(LL0)sin2kmt
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Solution

The correct option is A L0+(LL0)cos2kmt
In COM frame both the masses execute SHM with
ω=kμ

Here, μ= reduced mass
μ=m1×m2m1+m2=m2m+m=m2

ω=2km

Initially, particles are at extreme and distance between them is L. Hence, required distance =L0+(LL0)cos2kmt

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